A daring ranch hand sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed of the horse is 15.0 m/s, and the distance from the limb to the level of the saddle is 3.50 m.

(a) What must be the horizontal distance between the saddle and limb when the ranch hand makes his move?

(b) How long is he in the air?

I used the kinematic equations of motion but I can't seem to get an answer that makes sense.

He needs to fall 3.5m

time in air: 3.5=-1/2 g t^2
now having the time in air,
distance= 15*t

To solve for the first part, I got time = 2.89 seconds, but then it shows that I got the wrong answer?

Nah bobpursely is definitely wrong. Anyone know how to solve this?

Yea the solution provided is incorrect.

The correct solution is this:

3.5/4.9 = some answer s^2

then remember

someanswer s^2 = t^2

so solve for t to get .845 seconds for time in air.

To solve this problem, we can use the kinematic equations of motion. However, it's important to note that the horizontal and vertical motions are completely independent of each other. Let's break the problem down into two parts: finding the horizontal distance (a) and finding the time in the air (b).

(a) Finding the horizontal distance:

In this case, we can treat the horizontal motion as uniform and use the equation:

distance = speed × time

We know that the speed of the horse is 15.0 m/s, and we need to find the time it takes for the ranch hand to drop onto the horse. Since we're looking for the horizontal distance, we can ignore the vertical motion for now. Let's call the time it takes for the ranch hand to drop t.

The horizontal distance covered by the horse during time t is equal to the horizontal distance between the saddle and the limb. Therefore:

distance = speed × time
distance = 15.0 m/s × t

But we'll need to find the value of t later, so let's leave it as a variable for now.

(b) Finding the time in the air:

To find the time in the air, we need to look at the vertical motion. The key equation to use here is:

distance = initial velocity × time + (1/2) acceleration × time^2

In this case, the initial vertical velocity is 0 because the ranch hand is dropping vertically. The acceleration is due to gravity, and we'll assume it to be -9.8 m/s^2 (taking downwards as positive).

Using this equation, we can find the time it takes for the ranch hand to reach the saddle, which is also the time in the air:

3.50 m = 0 × t + (1/2)(-9.8 m/s^2)(t^2)

Now we have a quadratic equation. Simplifying it further:

3.50 m = (1/2)(-9.8 m/s^2)(t^2)
3.50 m = -4.9 m/s^2 × t^2

Rearranging the equation:

t^2 = (3.50 m) / (-4.9 m/s^2)

Taking the square root of both sides:

t = sqrt[(3.50 m) / (-4.9 m/s^2)]

Now, we can substitute this value of t into the equation we got in part (a) to find the horizontal distance:

distance = 15.0 m/s × t

Substituting t:

distance = 15.0 m/s × sqrt[(3.50 m) / (-4.9 m/s^2)]

It's important to note that you may get a negative value under the square root. Since time cannot be negative, we discard this negative solution and take the positive square root.

Now, you can plug in the values into the equation to find the horizontal distance covered by the horse when the ranch hand makes his move.